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The intensity at the maximum in a Young’s double slit experiment is I₀. Distance between two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d?

1. I₀
2. I₀/4
3. 3/4 I₀
4. I₀/2

Correct answer: I₀/2

Solution

In front of one of the slits, the path difference between the two waves is λ/2, leading to destructive interference. The intensity at this point is given by I = I₀/4.

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